Logaritm Question!!!

- anonymous

Logaritm Question!!!

- chestercat

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- anonymous

Find a relation between x and y that does NOT involve logarithms:
log x + log y = log (x+ y)

- zepdrix

Can we assume it's a log of base 10?

- anonymous

yes i think so

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## More answers

- anonymous

logx + log y is never log(x+y) it seems... it is log(xy)

- anonymous

do you want the answer?

- zepdrix

I suppose we could exponentiate both sides..
\[\huge 10^{(\log x + \log y)}=10^{\log(x+y)}\]
Using rules of exponents we can split up the left side like this:
\[\huge 10^{\log x}*10^{\log y}=10^{\log(x+y)}\]
Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". Giving us:
\[\huge xy=(x+y)\]
\[\huge y=\frac{x}{x-1}\]

- zepdrix

Something like that maybe? :o

- anonymous

@zepdrix Oooh, sexy!

- zepdrix

lol :3

- anonymous

@zeptr
yes thats the answer !!!

- zepdrix

Oh cool. :)

- zepdrix

Were you just testing us or something? :D lol
Knew the answer the whole time huh? XD

- zepdrix

@AERONIK , it seems that logx+logy=log(x+y) sometimes.
I wouldn't say NEVER.
Obviously it doesn't match the rule for logs that we had in mind.
But if you plug in y=2, x=2 into the original equation, that should work out ok as one solution :D Since 2+2 is the same as 2*2

- anonymous

no i had the answers at the back of book but i have no idea how to do it... i still dont get the logic though...

- anonymous

so it is not an identity and just a function you say ?

- anonymous

yeh just a random equation

- zepdrix

Yah it appears it wasn't testing you on whether or not you knew your log identities, they just wanted to see a relationship between x and y. Interesting problem :O

- anonymous

and one more thing in your proof you have assumed the given statement to be correct, can you please give a formal proof for te same?

- zepdrix

what now? :o

- anonymous

and is it log to the base 10 or log to the base e ?

- zepdrix

For this problem, we have to assume that they are ALL the same base. Otherwise it won't work out the same. If they're a different base, let's say A, then we exponentiate, writing each side with a base A (regardless of what A might be) and it will work out the same! :D

- zepdrix

Different than 10 i mean*

- anonymous

oh ok. sorry ..

- anonymous

10 logx âˆ—10 logy =10 log(x+y) Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". HOW DID YOU DO THIS STEP... CAN YOU SHOW ME? PLEASE

- anonymous

by defination if y=log x to the base a, then x is nothing but a^y, just use this and you can prove the identity.

- zepdrix

\[\large \arcsin( \sin x)=x\]\[\huge e^{\ln x}=x\]
Yah this is true of any function and it's inverse :D Remember it in trig with the inverse functions?

- anonymous

no ...can you do a proof for it please?

- zepdrix

|dw:1353225574033:dw|
Hmmm this might be a little confusing.. I drew a lot of arrows :| lemme know if it makes some sense.

- anonymous

i get it ...thanks so much

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